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Magazine Chapter 31: Problem 18
Solve the given differential equations. $$2 \frac{d^{2} y}{d x^{2}}+11 \frac{d y}{d x}-6 y=8 x$$
Short Answer
Expert verified
The general solution is \( y = C_1 e^{\frac{1}{2} x} + C_2 e^{-3x} - \frac{4}{3}x - \frac{22}{9} \).
Step by step solution
01
Identify the Type of Differential Equation
This is a non-homogeneous second-order linear differential equation with constant coefficients. It can be written in the general form as: \( a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + c y = g(x) \) where \( a=2, b=11, c=-6, \text{ and } g(x)=8x \).
02
Solve the Associated Homogeneous Equation
Consider the associated homogeneous equation: \( 2 \frac{d^2 y}{d x^2} + 11 \frac{d y}{d x} - 6 y = 0 \). To solve, find the characteristic equation: \( 2m^2 + 11m - 6 = 0 \).
03
Solve the Characteristic Equation
Solve the quadratic equation \( 2m^2 + 11m - 6 = 0 \) using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). Substituting, we get:\[ m = \frac{-11 \pm \sqrt{121 + 48}}{4} = \frac{-11 \pm \sqrt{169}}{4} = \frac{-11 \pm 13}{4} \]Thus, the roots are \( m_1 = \frac{1}{2} \) and \( m_2 = -3 \).
04
Write the General Solution of the Homogeneous Equation
Since the roots \( m_1 \) and \( m_2 \) are distinct, the general solution of the homogeneous equation is:\( y_h = C_1 e^{m_1 x} + C_2 e^{m_2 x} \)Substituting the roots, we get:\( y_h = C_1 e^{\frac{1}{2} x} + C_2 e^{-3x} \), where \( C_1 \) and \( C_2 \) are arbitrary constants.
05
Find a Particular Solution
To find a particular solution \( y_p \) for the non-homogeneous equation, we assume a form similar to \( g(x) \). Here, a linear function like \( Ax + B \) fits. We differentiate:\( \frac{dy_p}{dx} = A \), \( \frac{d^2y_p}{dx^2} = 0 \).Substitute into the differential equation:\[ 2(0) + 11(A) - 6(Ax + B) = 8x \]Simplify and equate coefficients:\[ 11A - 6Ax - 6B = 8x \]Matching coefficients, \( -6A = 8 \) gives \( A = -\frac{4}{3} \), and \( 11A - 6B = 0 \) gives \( B = -\frac{22}{9} \). Thus, \( y_p = -\frac{4}{3}x - \frac{22}{9} \).
06
Combine to Form the General Solution
The general solution of the non-homogeneous equation is:\( y = y_h + y_p \)Thus, \[ y = C_1 e^{\frac{1}{2} x} + C_2 e^{-3x} - \frac{4}{3}x - \frac{22}{9} \], where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
non-homogeneous differential equations
A non-homogeneous differential equation is one where the right-hand side is not equal to zero. In more precise terms, these equations can be expressed in the form:\[ a \frac{d^2 y}{d x^2} + b \frac{d y}{d x} + c y = g(x) \]Here, \( g(x) \) is a non-zero function that depends on \( x \). This contrasts with homogeneous differential equations where \( g(x) = 0 \).
Non-homogeneous differential equations are quite common in physics and engineering because they describe systems influenced by external forces or inputs. For instance, in our original exercise, \( g(x) = 8x \) represents such an external influence.
Non-homogeneous differential equations are quite common in physics and engineering because they describe systems influenced by external forces or inputs. For instance, in our original exercise, \( g(x) = 8x \) represents such an external influence.
- These equations often require finding both a homogeneous solution and a specific particular solution.
- The solution to the equation can typically be expressed as a sum of these two parts.
characteristic equation
The characteristic equation is a crucial concept when solving linear differential equations, particularly those with constant coefficients. It is derived from the associated homogeneous differential equation:\[ a \frac{d^2 y}{d x^2} + b \frac{d y}{d x} + c y = 0 \]In our exercise, we solved the homogeneous form:\[ 2 \frac{d^2 y}{d x^2} + 11 \frac{d y}{d x} - 6 y = 0 \]To tackle this, transform it into the characteristic equation by substituting \( y = e^{mx} \):\[ 2m^2 + 11m - 6 = 0 \]This is a quadratic equation in terms of \( m \) and its solutions give crucial insights:
- The roots of the characteristic equation determine the form of the general solution of the homogeneous part.
- Distinct real roots yield exponential solutions, while repeated roots or complex roots require adjustments to the solution form.
general solution
The general solution of a differential equation is a comprehensive expression that encompasses the set of all possible solutions. When dealing with non-homogeneous differential equations, the general solution is composed of two parts:
- \( y_h \): The general solution of the associated homogeneous equation.
- \( y_p \): A particular solution of the non-homogeneous equation.
- The homogeneous solution \( y_h \) is given by the expression \( C_1 e^{\frac{1}{2} x} + C_2 e^{-3x} \), where \( C_1 \) and \( C_2 \) are constants based on initial or boundary conditions.
- The particular solution \( y_p \) accounts for the non-zero right-hand side, ensuring the solution fits the entire original equation.
particular solution
When solving non-homogeneous differential equations, finding a particular solution is essential. This solution, \( y_p \), is a specific function that satisfies the non-homogeneous equation under the influence of \( g(x) \).
- In practice, one typically assumes a form for \( y_p \) that mirrors the type of function in \( g(x) \).
- For a polynomial \( g(x) \), which is our case (\( 8x \)), you might try a polynomial form like \( Ax + B \).
- Solve for coefficients by aligning like terms on both sides.
- In our case, solving gives: \( A = -\frac{4}{3} \) and \( B = -\frac{22}{9} \).
